Does a Continuous Function Need to Be One-to-One?

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In summary: Were you lucky enough to redo your proof?I did later submit a proof on paper to the professor but that day I just shrank into a small lump in my chair. The next time I was called on to present a proof in class I did well and I actually passed the course!
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JasonRox
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I have a question regarding this.

I wish I were home right now so I can give the exact words.

Anyways, the book is talking about continuous maps from one space to another. This is basically what it says...

Let f be a function with domain D in R. Then the following statements are equivalent:
f is continuous
If D is open, then the inverse image of every open set under f is again open.
If D is closed, then the inverse image of every closed set under f is again closed.

There are others, but that's not important.

I just want to clarify that f does not need to be one-to-one, correct?
 
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No, f does not need to be one to one. Instead of 'inverse image' try using the phrase 'preimage' of a set as someone here once pointed out. Can't remember who to credit it with.
 
  • #3
matt grime said:
No, f does not need to be one to one. Instead of 'inverse image' try using the phrase 'preimage' of a set as someone here once pointed out. Can't remember who to credit it with.

So, using the "preimage" we can "map" to two elements?

I used quotes for map because it doesn't really satisfy the definition.
 
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  • #4
I'm not sure what you mean by "map" two elements but here is an example:

If f(x)= x2 (not one-to-one) and D is the interval (-1, 4) then f-1(D) = (-2, 2), the set of numbers whose square is between -1 and 4. Since (-1, 4) is open and f is continuous, that inverse image (preimage if you prefer) is open.
 
  • #5
HallsofIvy said:
I'm not sure what you mean by "map" two elements but here is an example:

If f(x)= x2 (not one-to-one) and D is the interval (-1, 4) then f-1(D) = (-2, 2), the set of numbers whose square is between -1 and 4. Since (-1, 4) is open and f is continuous, that inverse image (preimage if you prefer) is open.

That's exactly what I wanted to confirm. Thanks.
 
  • #6
My most embarassing moment: My first semester in graduate school, I was called on to do a proof in Topology class that involved f-1(A) where A is a set. Without thinking, I did the proof assuming that f had an inverse! (Hey, it said f-1!)
 
  • #7
HallsofIvy said:
My most embarassing moment: My first semester in graduate school, I was called on to do a proof in Topology class that involved f-1(A) where A is a set. Without thinking, I did the proof assuming that f had an inverse! (Hey, it said f-1!)

I'll remember this whenever I get embarrased. :wink:

Were you lucky enough to redo your proof?
 
  • #8
I did later submit a proof on paper to the professor but that day I just shrank into a small lump in my chair. The next time I was called on to present a proof in class I did well and I actually passed the course!
 

FAQ: Does a Continuous Function Need to Be One-to-One?

What is a continuous mapping?

A continuous mapping, also known as a continuous function, is a mathematical concept that describes a mapping between two sets where small changes in the input result in small changes in the output. In simpler terms, it means that the graph of the function has no sudden breaks or gaps.

How do you determine if a mapping is continuous?

A mapping can be determined to be continuous if it satisfies the epsilon-delta definition of continuity. This means that for any epsilon (small positive number) chosen, there exists a delta (small positive number) such that the distance between the input and the output of the function is less than epsilon for all values within delta of the input. In short, the function must have no sudden jumps or breaks and must be able to be drawn without lifting the pencil from the paper.

What is the importance of continuous mappings in mathematics?

Continuous mappings are important in mathematics as they help us understand the behavior of functions and their graphs. They allow us to make predictions and solve problems in various fields such as physics, engineering, economics, and more. Additionally, many mathematical concepts, such as limits and derivatives, rely on the idea of continuity.

Can a function be continuous at one point but not at others?

Yes, a function can be continuous at one point and not at others. This is known as local continuity. It means that the function is continuous at a specific point, but there may be other points where the function is not continuous. In order for a function to be continuous at all points, it must be continuous at every point in its domain.

How are continuous mappings related to differentiability?

Continuous mappings are related to differentiability in that a function must be continuous in order to be differentiable. This means that if a function is not continuous at a certain point, it cannot be differentiable at that point. However, a function can be continuous at a point without being differentiable at that point. In order for a function to be differentiable at a point, it must be both continuous and have a defined derivative at that point.

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